Which determines the reactivity of an alkali metal?

Chemistry

2 answers:

marissa [1.9K]3 years ago

8 0

Answer: D— the ability to lose electrons

Explanation:

Just took the quiz

adelina 88 [10]3 years ago

3 0

Answer:

its ability to lose electron

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In the formation of potassium chloride, the product, KCl, has less energy than the reactants, K and Cl2. This reaction is .

k0ka [10]

if the product has lesser energy than the reactants it means enthalpy change value is positive.so it is endothermic reaction where the heat is absorbed.so kcl will formation will have positive enthalpy change .

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3 years ago

Describe how chemists use mass as a way of getting a precise estimate of the number of entities in a sample.

jok3333 [9.3K]

Answer:

Explanation:

To start with, the mass of a substance is centered on the amount of particles contained in it.

Mass of a substance measure is the same anywhere it is taken.

In chemistry, the concept of mole is used to find the exact estimate of the number of entities in a sample.

A mole of substance is quantity that is used to express the number of particles contained in it.

1 mole of any substance contains the Avogadro's number of particles.

This amount of particles is given as 6.02 x 10²³; the particles can be protons, neutrons, electrons e.t.c.

To find the number of moles;

Number of moles = $\frac{mass}{molar mass}$

Through this, we clearly ascertain the number of entities in a sample given the mass.

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3 years ago

The radioactive atom 61/27 co emits a beta particle. write an equation showing the decay

pickupchik [31]

Answer:

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{Ni}$

Explanation:

The unbalanced nuclear equation is

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, ?$

Let's write the question mark as a nuclear symbol.

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{Z}^{A}\text{X}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.

Then

60 = 0 + A, so A = 60 - 0 = 60, and

27 = -1 + Z, so Z = 27 + 1 = 28

Your nuclear equation becomes

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{X}$